Fair Coin Toss?

Discussion in 'Probability' started by nstyrthy, Sep 15, 2008.

  1. nstyrthy

    nstyrthy Guest

    Assuming we flip a fair coin 9 times and get 9 heads. What's the
    probability that the coin will come up heads on the next toss?

    I'm a li'l confused here. They say a fair coin toss is independent of
    other toss, and so the probability should be 50/50. But per Bayes
    theorem, we need to revise the probability and that means there's 9/10
    chance of getting heads on the next flip (assuming my calculation is
    right). But then again, statistics show that of 10 flips, roughly 5
    should be heads, and that means, this fair coin is *overdue* for
    showing tails, and so has a high probability for coming up tails.

    Clearly, I'm wrong somewhere. What am I missing?

    Any help will be greatly appreciated. Thanks.
    nstyrthy, Sep 15, 2008
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  2. nstyrthy

    se16 Guest

    You have said it is a fair coin. So by definition the chance should
    be 50% every time.

    Your Bayesian analysis would produce the same result, provided that
    before the execise started you were absolutely certain that you had a
    fair coin (known as a point prior). If however you had been prepared
    to entertain some doubt then the chance that next flip would be heads
    would be more than 50%. How much more would depend on the nature and
    extent of your doubt: for example, if you had thought that there had
    been a 0.5% chance of a double-headed coin, a 0.5% chance of a double
    tailed coin and a 99% chance of a fair coin then after nine heads you
    would revise these to about 72%, 28% and 0% repectively, so your new
    belief would be that the next coin had about an 86% chance of being

    Your "overdue" thinking contradicts the memoryless nature of coin
    flips. You could use such an argument if for example you were taking
    balls out of a bag and not putting them back, where you knew that
    originally there were equal numbers of red and yellow balls (and you
    could calculate a probability if you knew how many there were).
    se16, Sep 15, 2008
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  3. nstyrthy

    illywhacker Guest

    Hi. Sorry: I do not know what to call you as you have not signed your

    You are indeed confusing three different things.

    1) The knowledge you have (i.e. the probability you assign) *before*
    see the coin tossed, assuming have good reason to believe that the
    tossing method is 'fair'. This is the situation in which you assign
    probability p = 0.5 to getting heads on each coin toss and treat the
    as independent. In this case:

    - The best prediction for the number of heads you will get in ten
    tosses is

    - The outcomes of the tosses are independent because knowing the
    outcome of
    one toss will not affect the probability you assign to the outcome of
    another, since you assume you know that p = 0.5.

    2) The knowledge you have (i.e. the probability you assign) *before*
    see the coin tossed, assuming now that you have *no* particular
    about the coin or the coin tossing method, i.e. you do not know if it
    'fair' or not. In this situation, you do not know what probability to
    assign to each coin toss, so you have to sum equally over all
    for p. In this case:

    - The best prediction for the number of heads you will get in ten
    tosses is
    still five.

    - But the outcomes of the tosses are not independent because knowing
    outcome of one toss would help you to estimate the probability p of
    heads on another toss and hence influence your prediction for the

    3) The knowledge you have (i.e. the probability you assign) *after*
    have seen the coin come up heads ten times. What you know now
    depends on what you knew before you saw the tosses, as another poster
    indicated. If you were previously very certain that p = 0.5, i.e. case
    then this might be revised a little, but not much. You will put down
    ten heads to a 'fluke'. However, if you were previously completely
    about p, as in case (2), your uncertainty about p has now been
    reduced, and
    what is more you have good reason to doubt that the coin tossing
    method is

    Thus this case is intermediate between (1) and (2). In (1), you assume
    are certain about p. In (2), you assume you know nothing about p, i.e.
    are maximally uncertain. In (3), even if you start off knowing nothing
    about p, the results of the tosses tell you something, although even
    so you
    cannot be certain. Hence:

    - Your best prediction of the number of heads you will get in the next
    tosses is no longer five, since the results indicate that p is very
    unlikely to be 0.5.

    - The residual uncertainty means that seeing the results of more
    tosses can
    still tell you more about p, so the tosses are still not completely
    independent, although more so than in (2).


    The *most important point* here is this: these are *not* statements
    the coin, the coin tossing method, physics, or any fact to do with the
    world. They are statements about your *knowledge* of these things. If
    do not get to grips with this idea, you will be coming back here again
    again asking the same questions that many others have asked, confused
    their belief that probability is a property of the world, like weight,
    rather than a description of our knowledge of the world.

    Luckily this idea is very intuitive, so intuitive in fact that people
    taking statistics courses have to have it beaten out of them by
    intellectual intimidation, and even then can barely stop themselves
    using the forms of language to which it gives rise, e.g. 'the
    of the parameter value'. See any of the many discussions concerning
    'how to
    talk about confidence intervals' to get an idea of the tortuous nature
    the alternatives.

    illywhacker, Sep 15, 2008
  4. nstyrthy

    nstyrthy Guest

    Thanks everyone for the help.

    illywhacker, thanks for the very clear and lucid explanation. T
    nstyrthy, Sep 16, 2008
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